a/

\(y=1-sin^2x+2sinx+2=4-\left(sinx-1\right)^2\le4\)

\(y_{max}=4\) khi \(sinx=1\)

Mặt khác \(sinx\ge-1\Rightarrow\left(sinx-1\right)^2\le4\)

\(y_{min}=4-4=0\) khi \(sinx=-1\)

b/

\(y=sin^4x-2\left(1-sin^2x\right)+1=sin^4x+2sin^2x-1\)

Do \(0\le sin^2x\le1\)

\(\Rightarrow-1\le y\le2\)

\(y_{min}=-1\) khi \(sinx=0\)

\(y_{max}=2\) khi \(sin^2x=1\)

c/

\(y=2\left(1-cos2x\right)+sin2x+cos2x\)

\(=sin2x-cos2x+2=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)+2\)

Do \(-1\le sin\left(2x-\frac{\pi}{4}\right)\le1\)

\(\Rightarrow2-\sqrt{2}\le y\le2+\sqrt{2}\)

\(y_{min}=2-\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=-1\)

\(y_{max}=2+\sqrt{2}\) khi \(sin\left(2x+\frac{\pi}{4}\right)=1\)

d/

\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)

\(=1-3sin^2x.cos^2x\)

\(=1-\frac{3}{4}sin^22x\)

Mà \(0\le sin^22x\le1\Rightarrow\frac{1}{4}\le y\le1\)

\(y_{min}=\frac{1}{4}\) khi \(sin^22x=1\)

\(y_{max}=1\) khi \(sin2x=0\)

e/

\(y=5sinx+6cosx-7\)

\(=\sqrt{61}\left(\frac{5}{\sqrt{61}}sinx+\frac{6}{\sqrt{61}}cosx\right)-7\)

\(=\sqrt{61}\left(sinx.cosa+cosx.sina\right)-7\) (với \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{5}{\sqrt{61}}\))

\(=\sqrt{61}.sin\left(x+a\right)-7\)

Do \(-1\le sin\left(x+a\right)\le1\Rightarrow7-\sqrt{61}\le y\le7+\sqrt{61}\)

\(y_{min}=7-\sqrt{61}\) khi \(sin\left(x+a\right)=-1\)

\(y_{max}=7+\sqrt{61}\) khi \(sin\left(x+a\right)=1\)

f/

\(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+3\)

\(=2sin\left(x+\frac{\pi}{3}\right)+3\)

\(\Rightarrow1\le y\le5\)

\(y_{min}=1\) khi \(sin\left(x+\frac{\pi}{3}\right)=-1\)

\(y_{max}=5\) khi \(x+\frac{\pi}{3}=1\)

a.

\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)

\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)

\(\Rightarrow y_{min}=y\left(1\right)=0\)

\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)

b.

\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]

\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)

c.

\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)

Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=-t^2-t+2\)

\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)

\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)

\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)

d.

Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)

\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)

\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)

a/

\(cos^6x+sin^2x=1\)

\(\Leftrightarrow cos^6x-\left(1-sin^2x\right)=0\)

\(\Leftrightarrow cos^6x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left(cos^4x-1\right)=0\)

\(\Leftrightarrow cos^2x\left(cos^2x-1\right)\left(cos^2x+1\right)=0\)

\(\Leftrightarrow-cos^2x.sin^2x=0\)

\(\Leftrightarrow sin^22x=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

b/

\(cos^6x-sin^6x=\frac{13}{18}cos^22x\)

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos^4x+sin^4x+sin^2x.cos^2x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left[\left(sin^2x+cos^2x\right)^2-sin^2x.cos^2x\right]=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(1-\frac{1}{4}sin^22x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(1-\frac{1}{4}\left(1-cos^22x\right)\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(\frac{3}{4}+\frac{1}{4}cos^22x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(\frac{1}{4}cos^22x-\frac{13}{18}cos2x+\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\frac{1}{4}cos^22x-\frac{13}{18}cos2x+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

c/

\(cos^4x+sin^6x=cos2x\)

\(\Leftrightarrow\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow cos^32x-5cos^2x+7cos2x-3=0\)

\(\Leftrightarrow\left(cos2x-1\right)^2\left(cos2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=k2\pi\)

\(\Rightarrow x=k\pi\)

a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)

b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx

⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x

⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x

⇔ 4sin2x + (sinx + cosx) . sin2x = 0

⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)

⇔ sin2x = 0

c, 2cos³x = sin3x

⇔ 2cos³x = 3sinx - 4sin³x

⇔ 4sin³x + 2cos³x - 3sinx(sin²x + cos²x) = 0

⇔ sin³x + 2cos³x - 3sinx.cos²x = 0

Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình

Xét cosx ≠ 0 chia cả 2 vế cho cos³x ta được : 

tan³x + 2 - 3tanx = 0

⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)

d, cos²x - \(\sqrt{3}sin2x\) = 1 + sin²x

⇔ cos²x - sin²x - \(\sqrt{3}sin2x\) = 1

⇔ cos2x - \(\sqrt{3}sin2x\) = 1

⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)

⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)

e, cos³x + sin³x = 2cos⁵x + 2sin⁵x

⇔ cos³x (1 - 2cos²x) + sin³x (1 - 2sin²x) = 0

⇔ cos³x . (- cos2x) + sin³x . cos2x = 0

⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)

\(a=\int\dfrac{1}{2tan^2x+5tanx+2}.\dfrac{dx}{cos^2x}\)

Đặt \(tanx=t\Rightarrow dt=\dfrac{dx}{cos^2x}\)

\(I=\int\dfrac{dt}{2t^2+5t+2}=\int\dfrac{dt}{\left(t+2\right)\left(2t+1\right)}=\dfrac{2}{3}\int\left(\dfrac{1}{2t+1}-\dfrac{1}{2t+4}\right)dt\)

\(=\dfrac{1}{3}ln\left|\dfrac{2t+1}{2t+4}\right|+C=\dfrac{1}{3}ln\left|\dfrac{2tanx+1}{2tanx+4}\right|+C\)

Câu b hoàn toàn tương tự

a: ĐKXĐ: \(x\in R\)

=>TXĐ: D=R

b; ĐKXĐ: 2x-4>=0

=>x>=2

TXĐ: D=[2;+\(\infty\))

c: ĐKXĐ: 1-cos^2x>=0

=>sin^2x>=0(luôn đúng)

a.

Tìm min:

$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$

Vậy $y_{\min}=2$

----------------

Mặt khác: 

$y=4\sin x(\sin x+1)-8(\sin x+1)+11$

$=(\sin x+1)(4\sin x-8)+11$

$=4(\sin x+1)(\sin x-2)+11$

Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$

$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$

$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$

Vậy $y_{\max}=11$

b.

$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$

$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$

Vậy $y_{\max}=4$.

---------------------------

Mặt khác:

$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$

$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$

$=(1+\sin x)(3-\sin x)$

Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$

$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$

Vậy $y_{\min}=0$

c.

$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$

$=\sin ^4x+2\sin ^2x-1$

$=(\sin ^4x-1)+(2\sin ^2x-2)+2$

$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$

$=(\sin ^2x-1)(\sin ^2x+3)+2$

Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$

$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$

$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$

Vậy $y_{\max}=2$

------------------------------------------

$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$

Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$

$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$